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One more rectangular-shaped piece of metal siding needs to be cut to cover the exterior of a pole barn. The area of the piece is 30 ft2. The length is 1 less than 3 times the width. How wide should the metal piece be? Round to the nearest hundredth of a foot.

 Jul 2, 2019
 #1
avatar+714 
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Let's call the width "x."

The width is x and the length is 3x - 1

The area of the width and length is 30.

(x)(3x-1) = 30

3x^2-x = 30

3x^2-x-30 = 0

(3x+3)(x-10) = 0

Solve 3x+3=0 and x-10=0 and see which value of "x" is feasible. (hint: widths cannot be negative.)

You are very welcome!

:P

 Jul 2, 2019
 #2
avatar+101813 
+2

L * W  = Area....so...

x (3x - 1)  = 30

3x^2 - x = 30

3x^2 - x -30  = 0      factor as

(3x - 10)(x + 3)  = 0

 

Set each factor to 0  an solve for x  and we have that

 

3x - 10  = 0                     x + 3  = 0

3x = 10                           x  = -3   (reject)

x =10/3 

 

The width is   3(10/3) - 1   =   30/3  - 1   = 10 -1  =  9 ft

 

cool cool cool

 Jul 2, 2019

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